\(\Leftrightarrow\left\{{}\begin{matrix}x-y+3xy=3\\\left(x-y\right)^2+3xy=3\end{matrix}\right.\)
Trừ vế cho vế:
\(\Rightarrow\left(x-y\right)^2-\left(x-y\right)=0\)
\(\Leftrightarrow\left(x-y\right)\left(x-y-1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}y=x\\y=x-1\end{matrix}\right.\)
Thế vào \(x-y+3xy=3\)
TH1: \(y=x\Rightarrow x-x-3x^2=3\Rightarrow\left[{}\begin{matrix}x=1\Rightarrow y=1\\x=-1\Rightarrow y=-1\end{matrix}\right.\)
Th2: \(y=x-1\Rightarrow x-\left(x-1\right)+3x\left(x-1\right)=3\)
\(\Leftrightarrow3x^2-3x-2=0\Rightarrow\left[{}\begin{matrix}x=\dfrac{3+\sqrt{33}}{6}\Rightarrow y=\dfrac{-3+\sqrt{33}}{6}\\x=\dfrac{3-\sqrt{33}}{6}\Rightarrow y=\dfrac{-3-\sqrt{33}}{6}\end{matrix}\right.\)
